Chemical Kinetics Test

Result

Chemical Kinetics Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

The rate constant of a reaction at temperature 200 $$ \mathrm{K} $$ is10 times less than the rate constant at 400 $$ \mathrm{K} $$ . What is theactivation energy $$ \left(E_{\mathrm{a}}\right) $$ of the reaction $$ (R= $$ gas constant)?

A
1842.4$$ R $$

B
921.2$$ R $$

C
460.6$$ R $$

D
230.3$$ R $$

Solution

C₁ = rate constant at temperature 200 k

C₂ = rate constant at temperature 400 k

Given that

C₁ = C₂ /10

T₁ = lower temperature = 200 k

T₂ = higher temperature = 400 k

Q = activation energy

R = gas constant

The relation between activation energy , temperature and rate constants is given as

ln( C₂/C₁ ) = (Q/R) [(1/T₂) - (1/T₁)]

inserting the above values in the equation

ln( C₂/(C₂ /10) ) = - (Q/R) [(1/400) - (1/200)]

ln(10) ) = - (Q/R) [(1/400) - (1/200)]

2.303 = - (Q/R) [(1/400) - (1/200)]

Q/R = 921

Q = 921 R
Question 2
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For a reaction $$A \longrightarrow B+C$$ , it was found that at the end of 10 minutes from the start the total optical rotation of the system was $${ 50 }^{ o }$$ and when the reaction is complete, it was $${ 100}^{ o }$$. Assuming that only B and C are optically active and dextrorotatory. Calculate the rate constant of this first order reaction.

A
$$0.693 mi{ n }^{ -1 }$$

B
$$0.0693 se{ c }^{ -1 }$$

C
$$0.0693 { min }^{ -1 }$$

D
$$0.00693 { sec }^{ -1 }$$
Question 3
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From the above figure, the activation energy for the reverse reaction would be:

A
$$-120\ kJ{ mol }^{ -1 }$$

B
$$+152\ kJ{ mol }^{ -1 }$$

C
$$-120\ kJ{ mol }^{ -1 }$$

D
$$+1760\ kJ{ mol }^{ -1 }$$
Question 4
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for a zero order reaction-

A
$$t_{\frac{1}{2}}\propto a$$

B
$$t_{\frac{1}{2}}\propto \dfrac{1}{a}$$

C
$$t_{\frac{1}{2}}\propto a^2$$

D
$$t_{\frac{1}{2}}\propto \dfrac{1}{a^2}$$

Solution
Question 5
1 / -0

For a first order reaction velocity constant is $$K={ 10 }^{ -3 }{ s }^{ -1 }$$,Two third life for it would be

A
1100 s

B
2200 s

C
3300 s

D
4400 s
Question 6
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The half life period for catalytic decomposition of $$ A B_{3} $$ at $$ 50 \mathrm{mm} $$ is found to be $$4 hrs$$ and at $$100 \mathrm{mm} $$ it is $$2hrs$$. The order of reaction is -

A
$$3$$

B
$$1$$

C
$$2$$

D
$$0$$

Solution
Question 7
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Which one of the following formula represents a first-order reaction?

A
$$K=\dfrac { x }{ t } $$

B
$$K=\dfrac { 1 }{ 2t } \left[ \dfrac { 1 }{ \left( 1-x \right) ^{ 2 } } -\dfrac { 1 }{ { a }^{ 2 } } \right] $$

C
$$K=\dfrac { 2.303 }{ t } { log }_{ 10 }\dfrac { a }{ \left( a-x \right) } $$

D
$$K=\dfrac { 1 }{ t } \dfrac { x }{ \left( a-x \right) } $$
Question 8
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A graph plotted between log $$t_{50\%}$$ vs log of concentration is a straight line. What conclusion can you draw from the given graph?

A
$$n=1,\ { t }_{ 1/2 }=\dfrac { 1 }{ k-a } $$

B
$$n=2,\ { t }_{ 1/2 }=1/a$$

C
$$n=1,\ { t }_{ 1/2 }= 0.693/k$$

D
none of these
Question 9
1 / -0

According to molecular collision theory, the reaction is subjected to:

A
Number of molecular collisions of reactant

B
Number of collisions between reactants and activated complex

C
The collision rate between reactants and product molecules

D
Number of effective molecular collisions of reactants

Solution
Question 10
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The reactions having very high values of energy of activation are generally________

A
Very slow

B
Very fast

C
Spontaneous

D
Medium fast.